Question

A pipe is closed at one end. The speed of sound in air is \(330\,m\,s^{-1}\) and the length of the pipe is \(55\,cm\). Find its fundamental frequency.

• A. \(150\,Hz\)
• B. \(300\,Hz\)
• C. \(75\,Hz\)
• D. \(600\,Hz\)

Hint:

For a pipe closed at one end, the fundamental frequency is \[ f=\frac{v}{4L}. \] Only odd harmonics are present in such pipes.

Answer 1

The Correct Option is A

Concept: A pipe closed at one end supports only odd harmonics. For the fundamental mode of vibration, the closed end acts as a displacement node and the open end acts as a displacement antinode. The fundamental wavelength for a closed organ pipe is given by \[ \lambda = 4L \] and the corresponding frequency is \[ f=\frac{v}{\lambda} =\frac{v}{4L}. \]

Step 1: Write the given quantities. Speed of sound, \[ v=330\,m\,s^{-1} \] Length of pipe, \[ L=55\,cm \] Converting into SI unit, \[ L=0.55\,m \]

Step 2: Calculate the wavelength of the fundamental mode. For a pipe closed at one end, \[ \lambda =4L \] Substituting \(L=0.55\,m\), \[ \lambda =4\times0.55 \] \[ \lambda =2.2\,m \]

Step 3: Calculate the frequency. Using \[ f=\frac{v}{\lambda} \] we obtain \[ f=\frac{330}{2.2} \] \[ f=150\,Hz \]

Step 4: Final answer. Therefore, \[ \boxed{f=150\,Hz} \] Hence the correct option is \[ \boxed{(A)} \]